Hello, please take into account this: I'm a beginner 
I need to assess which of the two following specifications of a model is more suitable and explain it.
This is what I obtained running properly on R the tools I had:
FORWARD STEPWISE SELECTION:
LASSO REGRESSION:
This shows the degrees of freedom and the percentage of deviance inside the model.
fit.lasso
##
## Call: glmnet(x = x.train, y = y.train, family = "gaussian")
##
## Df %Dev Lambda
## [1,] 0 0.00000 1.135000
## [2,] 5 0.03258 1.034000
## [3,] 9 0.09986 0.942000
## [4,] 17 0.22030 0.858300
## [5,] 17 0.34360 0.782100
## [6,] 17 0.44590 0.712600
## [7,] 17 0.53090 0.649300
## [8,] 17 0.60150 0.591600
## [9,] 17 0.66000 0.539000
## [10,] 17 0.70870 0.491200
## [11,] 17 0.74900 0.447500
## [12,] 17 0.78260 0.407800
## [13,] 17 0.81040 0.371500
## [14,] 17 0.83350 0.338500
## [15,] 17 0.85270 0.308500
## [16,] 17 0.86860 0.281100
## [17,] 17 0.88180 0.256100
## [18,] 17 0.89280 0.233300
## [19,] 17 0.90190 0.212600
## [20,] 17 0.90950 0.193700
## [21,] 17 0.91580 0.176500
## [22,] 17 0.92100 0.160800
## [23,] 17 0.92530 0.146500
## [24,] 17 0.92890 0.133500
## [25,] 17 0.93190 0.121700
## [26,] 17 0.93440 0.110900
## [27,] 17 0.93640 0.101000
## [28,] 17 0.93810 0.092030
## [29,] 17 0.93950 0.083860
## [30,] 17 0.94070 0.076410
## [31,] 17 0.94170 0.069620
## [32,] 17 0.94250 0.063430
## [33,] 17 0.94320 0.057800
## [34,] 17 0.94370 0.052660
## [35,] 17 0.94420 0.047990
## [36,] 17 0.94460 0.043720
## [37,] 17 0.94490 0.039840
## [38,] 17 0.94520 0.036300
## [39,] 18 0.94540 0.033070
## [40,] 18 0.94560 0.030140
## [41,] 18 0.94580 0.027460
## [42,] 18 0.94590 0.025020
## [43,] 18 0.94600 0.022800
## [44,] 18 0.94610 0.020770
## [45,] 18 0.94620 0.018930
## [46,] 18 0.94620 0.017250
## [47,] 19 0.94630 0.015710
## [48,] 19 0.94630 0.014320
## [49,] 19 0.94640 0.013050
## [50,] 19 0.94640 0.011890
## [51,] 19 0.94640 0.010830
## [52,] 19 0.94650 0.009868
## [53,] 19 0.94650 0.008992
## [54,] 19 0.94650 0.008193
## [55,] 19 0.94650 0.007465
## [56,] 19 0.94650 0.006802
## [57,] 19 0.94650 0.006198
These are the minimum and maximum value of Lambda
cv.fit$lambda.min
## [1] 0.006801877
cv.fit$lambda.1se
## [1] 0.03983874
These functions are useful to obtain the values of the coefficients associated with the variables, when Lambda is at its minimum and at its maximum
coef(cv.fit, s="lambda.min")
## 21 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 0.036065536
## V1 0.980640994
## V2 1.026499043
## V3 0.993698752
## V4 0.985566688
## V5 0.951476162
## V6 0.994761436
## V7 1.033719423
## V8 1.024623825
## V9 1.006812111
## V10 1.020157370
## V11 1.012844843
## V12 1.009196928
## V13 1.019196655
## V14 1.023799094
## V15 0.970762354
## V16 0.993270801
## V17 0.953188089
## V18 -0.009839538
## V19 0.029612698
## V20 .
coef(cv.fit, s="lambda.1se")
## 21 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 0.04177626
## V1 0.94452267
## V2 0.99316613
## V3 0.95798195
## V4 0.94838097
## V5 0.92422061
## V6 0.96073814
## V7 1.00319146
## V8 0.98976810
## V9 0.97213574
## V10 0.98853023
## V11 0.97560603
## V12 0.97945382
## V13 0.98106038
## V14 0.98638349
## V15 0.93706507
## V16 0.95779553
## V17 0.91915907
## V18 .
## V19 .
## V20 .
In conclusion, I calculated the MSE of the lasso regression and of the full model, in order to check if the lasso method increased the reliability of our prediction:
y.test.lasso = predict(cv.fit, newx=x.test, s="lambda.min")
y.test.full = predict(cv.fit, newx=x.test, s=0)
pred.errors.lasso = y.test - y.test.lasso; # selected model predictions
pred.errors.full = y.test - y.test.full; # selected model predictions
mse.lasso = mean(pred.errors.lasso^2)
mse.full = mean(pred.errors.full^2)
mse.lasso
## [1] 0.9839925
mse.full
## [1] 0.983884
100 * (1- mse.lasso/mse.full) # percent gain in accuracy due to shrinkage
## [1] -0.01102601
If you arrived this far, THANK YOU. May I ask you to assess which one of the two cases is the most suitable?
I need to assess which of the two following specifications of a model is more suitable and explain it.
This is what I obtained running properly on R the tools I had:
FORWARD STEPWISE SELECTION:
LASSO REGRESSION:
This shows the degrees of freedom and the percentage of deviance inside the model.
fit.lasso
##
## Call: glmnet(x = x.train, y = y.train, family = "gaussian")
##
## Df %Dev Lambda
## [1,] 0 0.00000 1.135000
## [2,] 5 0.03258 1.034000
## [3,] 9 0.09986 0.942000
## [4,] 17 0.22030 0.858300
## [5,] 17 0.34360 0.782100
## [6,] 17 0.44590 0.712600
## [7,] 17 0.53090 0.649300
## [8,] 17 0.60150 0.591600
## [9,] 17 0.66000 0.539000
## [10,] 17 0.70870 0.491200
## [11,] 17 0.74900 0.447500
## [12,] 17 0.78260 0.407800
## [13,] 17 0.81040 0.371500
## [14,] 17 0.83350 0.338500
## [15,] 17 0.85270 0.308500
## [16,] 17 0.86860 0.281100
## [17,] 17 0.88180 0.256100
## [18,] 17 0.89280 0.233300
## [19,] 17 0.90190 0.212600
## [20,] 17 0.90950 0.193700
## [21,] 17 0.91580 0.176500
## [22,] 17 0.92100 0.160800
## [23,] 17 0.92530 0.146500
## [24,] 17 0.92890 0.133500
## [25,] 17 0.93190 0.121700
## [26,] 17 0.93440 0.110900
## [27,] 17 0.93640 0.101000
## [28,] 17 0.93810 0.092030
## [29,] 17 0.93950 0.083860
## [30,] 17 0.94070 0.076410
## [31,] 17 0.94170 0.069620
## [32,] 17 0.94250 0.063430
## [33,] 17 0.94320 0.057800
## [34,] 17 0.94370 0.052660
## [35,] 17 0.94420 0.047990
## [36,] 17 0.94460 0.043720
## [37,] 17 0.94490 0.039840
## [38,] 17 0.94520 0.036300
## [39,] 18 0.94540 0.033070
## [40,] 18 0.94560 0.030140
## [41,] 18 0.94580 0.027460
## [42,] 18 0.94590 0.025020
## [43,] 18 0.94600 0.022800
## [44,] 18 0.94610 0.020770
## [45,] 18 0.94620 0.018930
## [46,] 18 0.94620 0.017250
## [47,] 19 0.94630 0.015710
## [48,] 19 0.94630 0.014320
## [49,] 19 0.94640 0.013050
## [50,] 19 0.94640 0.011890
## [51,] 19 0.94640 0.010830
## [52,] 19 0.94650 0.009868
## [53,] 19 0.94650 0.008992
## [54,] 19 0.94650 0.008193
## [55,] 19 0.94650 0.007465
## [56,] 19 0.94650 0.006802
## [57,] 19 0.94650 0.006198
These are the minimum and maximum value of Lambda
cv.fit$lambda.min
## [1] 0.006801877
cv.fit$lambda.1se
## [1] 0.03983874
These functions are useful to obtain the values of the coefficients associated with the variables, when Lambda is at its minimum and at its maximum
coef(cv.fit, s="lambda.min")
## 21 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 0.036065536
## V1 0.980640994
## V2 1.026499043
## V3 0.993698752
## V4 0.985566688
## V5 0.951476162
## V6 0.994761436
## V7 1.033719423
## V8 1.024623825
## V9 1.006812111
## V10 1.020157370
## V11 1.012844843
## V12 1.009196928
## V13 1.019196655
## V14 1.023799094
## V15 0.970762354
## V16 0.993270801
## V17 0.953188089
## V18 -0.009839538
## V19 0.029612698
## V20 .
coef(cv.fit, s="lambda.1se")
## 21 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 0.04177626
## V1 0.94452267
## V2 0.99316613
## V3 0.95798195
## V4 0.94838097
## V5 0.92422061
## V6 0.96073814
## V7 1.00319146
## V8 0.98976810
## V9 0.97213574
## V10 0.98853023
## V11 0.97560603
## V12 0.97945382
## V13 0.98106038
## V14 0.98638349
## V15 0.93706507
## V16 0.95779553
## V17 0.91915907
## V18 .
## V19 .
## V20 .
In conclusion, I calculated the MSE of the lasso regression and of the full model, in order to check if the lasso method increased the reliability of our prediction:
y.test.lasso = predict(cv.fit, newx=x.test, s="lambda.min")
y.test.full = predict(cv.fit, newx=x.test, s=0)
pred.errors.lasso = y.test - y.test.lasso; # selected model predictions
pred.errors.full = y.test - y.test.full; # selected model predictions
mse.lasso = mean(pred.errors.lasso^2)
mse.full = mean(pred.errors.full^2)
mse.lasso
## [1] 0.9839925
mse.full
## [1] 0.983884
100 * (1- mse.lasso/mse.full) # percent gain in accuracy due to shrinkage
## [1] -0.01102601
If you arrived this far, THANK YOU. May I ask you to assess which one of the two cases is the most suitable?
